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10

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Probability arises when we perform an experiment that has various possible outcomes, but there is insuffi cient information to predict which of these outcomes will occur. The classic examples of this are tossing a coin, throwing a die, and drawing a card from a pack. Probability, however, is involved in almost every experiment done in science, and is fundamental to understanding statistics.

This chapter reviews some basic ideas of probability and develops a more systematic approach to solving probability problems. It concludes with the new topic of conditional probability.

Probability

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10A Probability and sample spaces 405

Probability and sample spaces

The first task is to develop a workable formula for probability that can serve as the foundation for the topic, we will do this when the possible results of an experiment can be divided into a finite number of equally likely possible outcomes.

Equally likely possible outcomesThe idea of equally likely possible outcomes is well illustrated by the experiment of throwing a die and recording the number shown. (A die, plural dice, is a cube with its corners and edges rounded so that it rolls easily, and with the numbers 1–6 printed on its six sides.) There are six possible outcomes: 1, 2, 3, 4, 5, 6. This is a complete list of the possible outcomes, because each time the die is rolled, one and only one of these outcomes will occur.

Provided that we believe the die to be completely symmetric and not biased in any way, there is no reason for us to expect that any one outcome is more likely to occur than any of the other five. We may thus regard these six possible outcomes as equally likely possible outcomes.

With the results of the experiment now divided into six equally likely possible outcomes, the probability 16 is

assigned to each of these six outcomes. Notice that these six probabilities are equal, and that they all add up to 1.

RandomnessThe explanations above assumed that the terms ‘equally likely’ and ‘more likely’ already have a meaning in the mind of the reader. There are many ways of interpreting these words. In the case of a thrown die, we could interpret the phrase ‘equally likely’ as a statement about the natural world, in this case that the die is perfectly symmetric. Alternatively, we could interpret it as saying that we lack entirely the knowledge to make any statement of preference for one outcome over another.

The word random can be used here. In the context of equally likely possible outcomes, saying that a die is thrown ‘randomly’ means that we are justified in assigning the same probability to each of the six possible outcomes. In a similar way, a card can be drawn ‘at random’ from a pack, or a queue of people can be formed in a ‘random order’.

10A

1 EQUALLY LIKELY POSSIBLE OUTCOMES

Suppose that the possible results of an experiment can be divided into a finite number n of equally likely possible outcomes — meaning that: • one and only one of these n outcomes will occur, and • we have no reason to expect one outcome to be more likely than another.

Then the probability 1n

is assigned to each of these equally likely possible outcomes.



Chapter 10 Probability406 10A

The basic formula for probabilitySuppose that a throw of at least 3 on a die is needed to win a game. Getting at least 3 on a die is called the particular event under discussion. The outcomes 3, 4, 5 and 6 are called favourable for this event, and the other two possible outcomes 1 and 2 are called unfavourable. The probability assigned to getting a score of at least 3 is then

P (scoring at least 3) = number of favourable outcomesnumber of possible outcomes

= 46

= 23

.

Again, this is easily generalised.

An experiment always includes what is recorded. For example, ‘tossing a coin and recording heads or tails’ is a different experiment from ‘tossing a coin and recording its distance from the wall’.

The sample space and the event spaceVenn diagrams and the language of sets are a great help when explaining probability or solving probability problems. Section 10C will present a short account of sets and Venn diagrams, but at this stage the diagrams should be self-explanatory.

The Venn diagram to the right shows the six equally likely possible outcomes when a die is thrown. The set of all these possible outcomes is

S = {1, 2, 3, 4, 5, 6}.

This set S is called a uniform sample space, and is represented by the outer rectangular box. The word ‘uniform’ indicates that the six possible outcomes are all equally likely, and the word ‘sample’ refers to the fact that running the experiment can be regarded as ‘sampling’. The event ‘scoring at least 3’ is identified with the set

E = {3, 4, 5, 6}

which is called the event space. It is represented by the ellipse, which is inside the rectangle because E is a subset of S.

Sample spaces and uniform sample spacesIn general, a sample space need not consist of equally likely possible outcomes. The only condition is that one and only one of the possible outcomes must occur when the experiment is performed.

A finite sample space is called uniform if it does consist of equally likely possible outcomes, as when a die is thrown.

2 THE BASIC FORMULA FOR PROBABILITY

If the results of an experiment can be divided into a finite number of equally likely possible outcomes, some of which are favourable for a particular event and the others unfavourable, then:

P (event) = number of favourable outcomesnumber of possible outcomes

E

S

12

34

5

6




The assumptions of the chapterFirst, the sample spaces in this chapter are assumed to be finite — this qualification applies without further mention throughout the chapter. Infinite sample spaces appear in the next chapter and in Year 12.

Secondly, the discussions are restricted to situations where the results of experiment can be reduced to a set of equally likely possible outcomes, that is, to a uniform sample space (assumed finite, as discussed above).

Thus, the basic probability formula can now be restated in set language as:

Probabilities involving playing cardsSo many questions in probability involve a pack of playing cards that anyone studying probability needs to be familiar with them. You are encouraged to acquire some cards and play some simple games with them. A pack of cards consists of 52 cards organised into four suits, each containing 13 cards. The four suits are:

two black suits: ♣ clubs ♠ spades

two red suits: ♦ diamonds ♥ hearts

Each of the four suits contains 13 cards:

A (ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king)

An ace is also regarded as a 1. Three cards in each suit are called court cards (or picture cards or face cards) because they depict people in the royal court:

J (Jack), Q (Queen), K (King)

It is assumed that when a pack of cards is shuffled, the order is totally random, meaning that there is no reason to expect any one ordering of the cards to be more likely to occur than any other.

4 THE BASIC FORMULA FOR PROBABILITY

Suppose that an event E is a subset of a uniform sample space S. Then

P (E) = ∣E∣∣S∣

,

where the symbols ∣E∣ and ∣S∣ mean the number of members of E and S.

3 SAMPLE SPACE

A set of possible outcomes of an experiment is called a sample space if: • one and only one of these outcomes will occur.

Uniform sample spaceA finite sample space is called uniform if: • all its possible outcomes are equally likely.

Event spaceThe set of all favourable outcomes is called the event space.




Impossible and certain eventsParts f and g of Example 1 were intended to illustrate the probabilities of events that are impossible or certain.• Getting a green card is impossible because there are no green cards. Hence there are no favourable

outcomes, and the probability is 0.• Getting a red or black card is certain because all the cards are either red or black. Hence all possible

outcomes are favourable, and the probability is 1.• For the other five events, the probability lies between 0 and 1.

Example 1 10A

A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is:a the seven of hearts b any heartc any seven d any red carde any court card (that is, a jack, a queen, or a king) f any green cardg any red or black card.

SOLUTION

In each case, there are 52 equally likely possible outcomes.

a There is 1 seven of hearts, so P (7 ♥) = 152

.

b There are 13 hearts, so P (heart) = 1352

= 14

.

c There are 4 sevens, so P (seven) = 452

= 113

.

d There are 26 red cards, so P (red card) = 2652

= 12

.

e There are 12 court cards, so P (court card) = 1252

= 313

.

f No card is green, so P (green card) = 052

= 0 .

g All 52 cards are red or black, so P (red or black card) = 5252

= 1.




In Section 10C, the complement E of a set E will be defined to be the set of things in S but not in E. The notation E for complementary event is quite deliberately the same notation as that for the complement of a set.

Complementary events and the word ‘not’It is often easier to find the probability that an event does not occur than the probability that it does occur. The complement of an event E is the event ‘E does not occur’. It is written as E, or sometimes as E ′ or E

c.

Let S be a uniform sample space of the experiment. The complementary event E is represented by the region outside the circle in the Venn diagram to the right. Because ∣E ∣ = ∣S ∣ − ∣E ∣ , it follows that

P (E) = ∣E ∣∣S ∣

= ∣S ∣ − ∣E ∣∣S ∣

= ∣S ∣∣S ∣

− ∣E ∣∣S ∣

= 1 − P (E).

E

S

E

5 IMPOSSIBLE AND CERTAIN EVENTS

• An impossible event has probability zero. • A certain event has probability 1. • The probability of any other event E lies in the interval 0 < P (E) < 1.

6 COMPLEMENTARY EVENTS AND THE WORD 'NOT'

• The complement E of an event E is the event ‘E does not occur’, normally read as ‘not E’? • P (E) = 1 − P (E). • The symbols E ′ and E

c are also used for the complementary event. • Always consider using complementary events in problems, particularly when the word ‘not’

occurs.

Example 2 10A

A card is drawn at random from a pack of playing cards. Find the probability that it isa not a spadeb not a court card ( jack, queen or king)c neither a red 2 nor a black 6.




Example 3 10A

Comment on the validity of this argument.‘Brisbane is one of 16 League teams, so the probability that Brisbane wins the premiership is 1

16.’

SOLUTION

[Identifying the fallacy]The division into 16 possible outcomes is correct (assuming that a tie for first place is impossible), but no reason has been offered as to why each team is equally likely to win, so the argument is invalid.

[Offering a replacement argument]If a team is selected at random from the 16 teams, then the probability that it is the premiership-winning team is 1

16. Also, if someone knows nothing whatsoever about League, then for them the statement is correct.

But these are different experiments.

Note: It is difficult to give a satisfactory account of this situation, indeed the idea of exact probabilities seems to have no meaning. Those with knowledge of the game would have some idea of ranking the 16 teams in order from most likely to win to least likely to win. If there is an organised system of betting, one may, or may not, agree to take this as an indication of the community’s collective wisdom on Brisbane’s chance of winning, and nominate it as ‘the probability’.

Invalid argumentsArguments offered in probability theory can be invalid for all sorts of subtle reasons, and it is common for a question to ask for comment on a given argument. It is most important in such a situation that any fallacy in the given argument be explained — it is not sufficient to offer only an alternative argument with a different conclusion.

SOLUTION

a Thirteen cards are spades, so P (spade) = 1352

= 14

.

Hence, using complementary events, P (not a spade) = 1 − 1

4

= 34

.

b Twelve cards are court cards, so P (court card) = 1252

= 313

.

Hence, using complementary events, P (court card) = 1 − 3

13

= 1013

.

c There are two red 2 s and two black 6 s, so P (red 2 or black 6) = 452

= 113

.

Hence, using complementary events, P (neither) = 1 − 1

13

= 1213

.




Experimental probability — relative frequencyWhen a drawing pin is thrown, there are two possible outcomes, point-up and point-down. But these two outcomes are not equally likely, and there seems to be no way to analyse the results of the experiment into equally likely possible outcomes.

In the absence of a fancy argument from physics about rotating pins falling on a smooth surface, however, some estimate of probability can be gained by performing the experiment many times. We can then use the relative frequencies of the drawing pin landing pin-up and pin-down as estimates of the probabilities, as in the following example. (Relative frequencies may have been introduced in earlier years.) But an exact value is unobtainable by these methods; indeed, the very idea of an exact value may well have no meaning.

The questions in the following example could raise difficult issues beyond this course, but the intention here is only that the questions be answered briefly in a common-sense manner. Chapter 11 pursues these issues in more detail, particularly in Section 11D.

Example 4 10A

A drawing pin was thrown 400 times and fell point-up 362 times.a What were the relative frequencies of the drawing pin falling point-up, and of falling point-down?b What probability does this experiment suggest for the result ‘point-up’?c A machine later repeated the experiment 1 000 000 times, and the pin fell point-up 916 203 times.

Does this change your estimate in part b?

SOLUTION

a Relative frequency of point-up = 362400

= 0.905

Relative frequency of point-down = 38400

= 0.095

b These results suggest that P (point-up) ≑ 0.905, but with only 400 trials, there would be little confidence in this result past the second decimal place, or even the first decimal place, because different runs of the same experiment would be expected to differ by small numbers. The safest conclusion would be that P (point-up) ≑ 0.9.

c The new results suggest that the estimate of the probability can now be refined to P (point-up) ≑ 0.916 — we can now be reasonably sure that the rounding to 0.9 in part a gave a value that was too low. (Did the machine throw the pin in a random manner, whatever that may mean?)

Exercise 10A

1 A pupil has 3 tickets in the class raffle. If there were 60 tickets in the raffle and one ticket is drawn, find the probability that the pupil:a wins b does not win.

2 A coin is tossed. Write down the probability that it shows:a a head b a tailc either a head or a tail d neither a head nor a tail.

FOUNDATION




3 If a die is rolled, find the probability that the uppermost face is:a a three b an even numberc a number greater than four d a multiple of three.

4 A bag contains five red and seven blue marbles. If one marble is drawn from the bag at random, find the probability that it is:a red b blue c green.

5 A bag contains eight red balls, seven yellow balls and three green balls. A ball is selected at random. Find the probability that it is:a red b yellow or green c not yellow.

6 In a bag there are four red capsicums, three green capsicums, six red apples and five green apples. One item is chosen at random. Find the probability that it is:a green b red c an appled a capsicum e a red apple f a green capsicum.

7 A letter is chosen at random from the word TASMANIA. Find the probability that it is:a an A b a vowel c a letter of the word HOBART.

8 A letter is randomly selected from the 26 letters in the English alphabet. Find the probability that the letter is:a the letter S b a vowelc a consonant d the letter γe a C, D or E f one of the letters of the word MATHS.Note: The letter Y is normally classified as a consonant.

9 A student has a 22% chance of being chosen as a prefect. What is the chance that they will not be chosen as a prefect?

10 When breeding labradors, the probability of breeding a black dog is 37 .

a What is the probability of breeding a dog that is not black?b If you bred 56 dogs, how many would you expect not to be black?

11 A box containing a light bulb has a chance of 115

of holding a defective bulb.

a If 120 boxes were checked, how many would you expect to hold defective bulbs?b What is the probability that the box holds a bulb that works?

12 When a die is rolled, the theoretical probability of throwing a six is 16 , which is about 17%.

a How many sixes would you expect to get if a die is thrown 60 times?b In an experiment, a certain die was thrown 60 times and a six turned up 18 times.

i What was the relative frequency of throwing a six?ii What does this experiment suggest is the experimental probability of throwing a six?iii Comment on whether the die is biased or fair.

13 Every day the principal goes early to the tuckshop to buy a sandwich for his lunch. He knows that the canteen makes 400 sandwiches a day — equal numbers of four types: bacon-lettuce-tomato; chicken; ham and cheese; and Vegemite. He always chooses a sandwich at random.a What is his theoretical probability of obtaining a chicken sandwich?b Over 20 days, the principal found he had bought 8 chicken sandwiches. What was his relative

frequency, and therefore his experimental probability, of obtaining a chicken sandwich?c Comment on why your answers for the theoretical and experimental probability might differ.




DEVELOPMENT

14 A number is selected at random from the integers 1, 2, 3, …, 19, 20. Find the probability of choosing:a the number 4 b a number greater than 15 c an even numberd an odd number e a prime number f a square numberg a multiple of 4 h the number e i a rational number.

15 From a regular pack of 52 cards, one card is drawn at random. Find the probability that:a it is black b it is red c it is a kingd it is the jack of hearts e it is a club f it is a picture cardg it is a heart or a spade h it is a red 5 or a black 7 i it is less than a 4.

16 A book has 150 pages. The book is randomly opened at a page. Find the probability that the page number is:a greater than 140 b a multiple of 20 c an odd numberd a number less than 25 e either 72 or 111 f a three-digit number.

17 An integer x, where 1 ≤ x ≤ 200, is chosen at random. Determine the probability that it:a is divisible by 5 b is a multiple of 13 c has two digitsd is a square number e is greater than 172 f has three equal digits.

18 A bag contains three times as many yellow marbles as blue marbles, and no other colours. If a marble is chosen at random, find the probability that it is:a yellow b blue.

19 Fifty tagged fish were released into a dam known to contain fish. Later, a sample of 30 fish was netted from this dam, of which eight were found to be tagged. Estimate the total number of fish in the dam just prior to the sample of 30 being removed.

CHALLENGE

20 Comment on the following arguments. Identify precisely any fallacies in the arguments, and, if possible, give some indication of how to correct them.a ‘On every day of the year it either rains or it doesn’t. Therefore the chance that it will rain tomorrow is 1

2 .’

b ‘When the Sydney Swans play Hawthorn, Hawthorn wins, the Swans win or the game is a draw.

Therefore the probability that the next game between these two teams results in a draw is 13 .’

c ‘When answering a multiple-choice question for which there are four possible answers to each

question, the chance that Peter answers the question correctly is 14 .’

d ‘A bag contains a number of red, white and black beads. If you choose one bead at random from the

bag, the probability that it is black is 13 .’

e ‘Four players play in a knockout tennis tournament resulting in a single winner. A man with no knowledge of the game or of the players declares that one particular player will win his semi-final,

but lose the final. The probability that he is correct is 14 .’

21 A rectangular field is 60 metres long and 30 metres wide. A cow wanders randomly around the field. Find the probability that the cow is:a more than 10 metres from the edge of the field,b not more than 10 metres from a corner of the field.



Chapter 10 Probability414 10B

Sample space graphs and tree diagrams

Many experiments consist of several stages. For example, when a die is thrown twice, the two throws can be regarded as two separate stages of the one experiment. This section develops two approaches — graphing and tree diagrams — to investigating the sample space of a multi-stage experiment.

Graphing the sample spaceThe word ‘sample space’ is used in probability, rather than ‘sample set’, because the sample space of a multi-stage experiment takes on some of the characteristics of a space. In particular, the sample space of a two-stage experiment can be displayed on a two-dimensional graph, and the sample space of a three-stage experiment can be displayed in a three-dimensional graph.

The following example shows how a two-dimensional dot diagram can be used for calculations with the sample space of a die thrown twice.

10B

Example 5 10B

A die is thrown twice. Find the probability that:a the pair is a double b at least one number is a fourc both numbers are greater than four d both numbers are evene the sum of the two numbers is six f the sum is at most four.

SOLUTION

The horizontal axis in the diagram to the right represents the six possible outcomes of the first throw, and the vertical axis represents the six possible outcomes of the second throw. Each dot is an ordered pair, such as (3, 5), representing the first and second outcomes. The 36 dots therefore represent all 36 different possible outcomes of the two-stage experiment, all equally likely. This is the full sample space.Parts a–f can now be answered by counting the dots representing the various event spaces.

a There are 6 doubles, so b 11 pairs contain a 4, so

P (double) = 636

(count the dots)

= 16

.

P (at least one is a 4) = 1136

.

1

1

3

5

2

4

6

1stthrow

2nd throw

2 3 4 5 6

1

3

5

2

4

6

1stthrow

2nd throw

654321

1

3

5

2

4

6

1stthrow

2nd throw

654321



10B Sample space graphs and tree diagrams 415

Example 6 10B

A die is thrown twice.a Find the probability of failing to throw a double six.b Find the probability that the sum is not seven.

SOLUTION

The same sample space can be used as in the previous example.

a The double six is one outcome out of 36,

so P (double six) = 136

.

Hence P (not throwing a double six) = 1 − P (double six)

= 3536

.

b The graph shows that there are six outcomes giving a sum of seven,

so P (sum is seven) = 636

= 16.

Hence P (sum is not seven) = 1 − P (sum is seven)

= 1 − 16

= 56

. 1

3

5

2

4

6

1stthrow

2nd throw

654321

c 4 pairs consist only of 5 or 6, so P (both greater than 4) = 436

= 19

.

d 9 pairs have two even members, so P (both even) = 936

= 14

.

e 5 pairs have sum 6, so P (sum is 6) = 536

.

f 6 pairs have sum 2, 3 or 4, so P (sum at most 4) = 636

= 16

.

Complementary events in multi-stage experimentsThe following example continues the two-stage experiment in the previous example, but this time it involves working with complementary events.

7 GRAPHING THE SAMPLE SPACE OF A TWO-STAGE EXPERIMENT

• The horizontal axis represents the first stage of the experiment. • The vertical axis represents the second stage.




The meaning of ‘word’In the example above and throughout this chapter, ‘word’ simply means an arrangement of letters — the arrangement doesn’t have to have any meaning or be a word in the dictionary. Thus ‘word’ simply becomes a convenient device for discussing arrangements of things in particular orders.

Example 7 10B

A three-letter word is chosen in the following way. The first and last letters are chosen from the three vowels ‘A’, ‘O’ and ‘U’, with repetition not allowed, and the middle letter is chosen from ‘L’ and ‘M’. List the sample space, then find the probability that:a the word is ‘ALO’ b the letter ‘O’ does not occurc ‘M’ and ‘U’ do not both occur d the letters are in alphabetical order.

SOLUTION

The tree diagram to the right lists all 12 equally likely possible outcomes. The two vowels must be different, because repetition was not allowed.

a P (‘ALO’) = 112

b P (no‘O’) = 412

= 13

c P (not both ‘M’ and ‘U’) = 812

= 23

d P (alphabetical order) = 412

= 13

A

O

U

L

M

L

M

L

M

OUOUAUAUAOAO

ALOALUAMOAMUOLAOLUOMAOMUULAULOUMAUMO

1stletter

2ndletter

3rdletter OutcomeStart

Tree diagramsListing the sample space of a multi-stage experiment can be difficult, and the dot diagrams of the previous paragraph are hard to draw in more than two dimensions. Tree diagrams provide a very useful alternative way to display the sample space. Such diagrams have a column for each stage, plus an initial column labelled ‘Start’ and a final column listing the possible outcomes.

8 A TREE DIAGRAM OF A MULTI-STAGE EXPERIMENT

• The first column is headed ‘Start’. • The last column is headed ‘Output’, and lists all the paths through the tree. • Each other column represents all the equally likely possible outcomes in the particular stage of the

experiment.




Invalid argumentsThe following example illustrates another invalid argument in probability. As always, the solution first offers an explanation of the fallacy, before then offering an alternative argument with a different conclusion.

Example 8 10B

Comment on the validity of this argument. ‘When two coins are tossed together, there are three outcomes: two heads, two tails and one of each.

Hence the probability of getting one of each is 13 .’

SOLUTION

[Identifying the fallacy]The division of the results into the three given outcomes is correct, but no reason is offered as to why these outcomes are equally likely.

[Supplying the correct argument]On the right is a tree diagram of the sample space. It divides the results of the experiment into four equally likely possible outcomes. Two of these outcomes, HT and TH, are favourable to the event ‘one of each’, so

P (one of each) = 24

= 12 .

HH

HT

TH

TT

H

T

H

TT

H

1stcoin

2ndcoin OutcomeStart

Exercise 10B

1 A fair coin is tossed twice.a Use a tree diagram to list the four possible outcomes.b Hence find the probability that the two tosses result in:

i two headsii a head and a tailiii a head on the first toss and a tail on the second.

2 A coin is tossed and a die is thrown.a Use a tree diagram to list all 12 possible outcomes.b Hence find the probability of obtaining:

i a head and an even number ii a tail and a number greater than 4iii a tail and a number less than 4 iv a head and a prime number.

3 Two tiles are chosen at random, one after the other, from three Scrabble tiles T, O and E, and placed in order on the table.a List the six possible outcomes in the sample space.b Find the probability of choosing:

i two vowelsii a consonant then a voweliii a T.

FOUNDATION




4 From a group of four students, Anna, Bill, Charlie and David, two are chosen at random, one after the other, to be on the Student Representative Council.a List the 12 possible outcomes in the sample space.b Hence find the probability that:

i Bill and David are chosen ii Anna is choseniii Charlie is chosen but Bill is not iv neither Anna nor David is selectedv Bill is chosen first vi Charles is not chosen second.

5 From the digits 2, 3, 8 and 9, a two-digit number is formed in which no digit is repeated.a Use a tree diagram to list the possible outcomes.b If the number was formed at random, find the probability that it is:

i the number 82 ii a number greater than 39iii an even number iv a multiple of 3v a number ending in 2 vi a perfect square.

6 A captain and vice-captain of a cricket team are to be chosen from Amanda, Belinda, Carol, Dianne and Emma.a Use a tree diagram to list the possible pairings, noting that order is important.b If the choices were made at random, find the probability that:

i Carol is captain and Emma is vice-captainii Belinda is either captain or vice-captainiii Amanda is not selected for either positioniv Emma is vice-captain.

DEVELOPMENT

7 A coin is tossed three times. Draw a tree diagram to illustrate the possible outcomes. Then find the probability of obtaining:a three heads b a head and two tailsc at least two tails d at most one heade more heads than tails f a head on the second toss.

8 A green die and a red die are thrown simultaneously. List the set of 36 possible outcomes on a two-dimensional graph and hence find the probability of:a obtaining a 3 on the green die b obtaining a 4 on the red diec a double 5 d a total score of 7e a total score greater than 9 f an even number on both diceg at least one 2 h neither a 1 nor a 4 appearingi a 5 and a number greater than 3 j the same number on both dice.

9 Suppose that the births of boys and girls are equally likely.a In a family of two children, determine the probability that there are:

i two girls ii no girls iii one boy and one girl.b In a family of three children, determine the probability that there are:

i three boys ii two girls and one boy iii more boys than girls.




10 A coin is tossed four times. Find the probability of obtaining:a four heads b exactly three tailsc at least two heads d at most one heade two heads and two tails f more tails than heads.

11 A hand of five cards contains a 10, jack, queen, king and ace. From the hand, two cards are drawn in succession, the first card not being replaced before the second card is drawn. Find the probability that:a the ace is drawnb the king is not drawnc the queen is the second card drawn.

CHALLENGE

12 Three-digit numbers are formed from the digits 2, 5, 7 and 8, without repetition.a Use a tree diagram to list all the possible outcomes. How many outcomes are there?b Hence find the probability that the number is:

i greater than 528ii divisible by 3iii divisible by 13iv prime.

13 If a coin is tossed n times, where n > 1, find the probability of obtaining:a n headsb at least one head and at least one tail.



Chapter 10 Probability420 10C

Sets and Venn diagrams

This section is a brief account of sets and Venn diagrams for those who have not met these ideas already. The three key ideas needed in probability are the intersection of sets, the union of sets, and the complement of a set.

Logic is very close to the surface when we talk about sets and Venn diagrams. The three ideas of intersection, union and complement mentioned above correspond very precisely to the words ‘and’, ‘or’ and ‘not’.

Listing sets and describing setsA set is a collection of things called elements or members. When a set is specified, it needs to be made absolutely clear what things are its members. This can be done by listing the members inside curly brackets. For example,

S = {1, 3, 5, 7, 9},

which is read as ‘S is the set whose members are 1, 3, 5, 7 and 9’.

It can also be done by writing a description of the members inside curly brackets. For example,

T = {odd integers from 0 to 10},

which is read as ‘T is the set of odd integers from 0 to 10’.

Equal setsTwo sets are called equal if they have exactly the same members. Hence the sets S and T in the previous paragraph are equal, which is written as S = T . The order in which the members are written doesn’t matter at all, and neither does repetition. For example,

{1, 3, 5, 7, 9} = {3, 9, 7, 5, 1} = {5, 9, 1, 3, 7} = {1, 3, 1, 5, 1, 7, 9}.

The size of a setA set may be finite, like the set above of positive odd numbers less than 10, or infinite, like the set of all integers. Only finite sets are needed here.

If a set S is finite, then the symbol ∣S ∣ is used to mean the number of members of S. For example:If A = {5, 6, 7, 8, 9, 10}, then ∣A ∣ = 6.If B = {letters in the alphabet}, then ∣B ∣ = 26.If C = {12}, then ∣C ∣ = 1.If D = {odd numbers between 1.5 and 2.5}, then ∣D ∣ = 0.

The empty setThe last set D above is called the empty set, because it has no members at all. The usual symbol for the empty set is ∅ . There is only one empty set, because any two empty sets have exactly the same members (that is, none at all) and so are equal.

10C



10C Sets and Venn diagrams 421

Intersection and unionThere are two obvious ways of combining two sets A and B. The intersection A ∩ B of A and B is the set of everything belonging to A and B. The union A ∪ B of A and B is the set of everything belonging to A or B.

For example, if A = {0, 1, 2, 3} and B = {1, 3, 6}, then

A ∩ B = {1, 3}A ∪ B = {0, 1, 2, 3, 6} .

Two sets A and B are called disjoint if they have no elements in common, that is, if A ∩ B = ∅. For example, if A = {2, 4, 6, 8} and B = {1, 3, 5, 7}, then

A ∩ B = ∅,

so A and B are disjoint.

‘And’ means intersection, ‘or’ means unionThe important mathematical words ‘and’ and ‘or’ can be interpreted in terms of union and intersection:

A ∩ B = {elements that are in A and in B}

A ∪ B = {elements that are in A or in B}

Note: The word ‘or’ in mathematics always means ‘and/or’. Correspondingly, all the elements of A ∩ B are members of A ∪ B.

9 SET TERMINOLOGY AND NOTATION

• A set is a collection of elements. For example, S = {1, 3, 5, 7, 9} and T = {odd integers from 0 to 10}. • Two sets are called equal if they have exactly the same members. • The number of elements in a finite set S is written as ∣S ∣ . • The empty set is written as ∅ — there is only one empty set.

10 THE INTERSECTION AND UNION OF SETS

Intersection of sets • The intersection of two sets A and B is the set of elements in A and in B:

A ∩ B = {elements in A and in B}.

• The sets A and B are called disjoint when A ∩ B = ∅.

Union of sets • The union of A and B is the set of elements in A or in B:

A ∪ B = {elements in A or in B}.

• The word ‘or’ in mathematics always means ‘and/or’.




Subsets of setsA set A is called a subset of a set B if every member of A is a member of B. This relation is written as A ⊂ B. For example,

{teenagers in Australia} ⊂ {people in Australia} {2, 3, 4} ⊂ {1, 2, 3, 4, 5}

{vowels} ⊂ {letters in the alphabet}.Because of the way subsets have been defined, every set is a subset of itself. Also, the empty set is a subset of every set. For example,

{1, 3, 5} ⊂ {1, 3, 5}, and ∅ ⊂ {1, 3, 5}.

The universal set and the complement of a setA universal set is the set of everything under discussion in a particular situation. For example, if A = {1, 3, 5, 7, 9}, then possible universal sets could be the set of all positive integers less than 11, or the set of all integers, or even the set of all real numbers.

Once a universal set E is fixed, then the complement A of any set A is the set of all members of that universal set which are not in A. For example,

If A = {1, 3, 5, 7, 9} and E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},then A = {2, 4, 6, 8, 10}.

Every member of the universal set is either in A or in A, but never in both A and A. This means that

A ∩ A = ∅, the empty set, andA ∪ A = E, the universal set.

‘Not’ means complementAs mentioned in Section 10A, the important mathematical word ‘not’ can be interpreted in terms of the complementary set:

A = {members of E that are not members of A}

The notations A′ and Ac are also used for the complement of A.

11 SUBSETS, THE UNIVERSAL SET, AND COMPLEMENTS

• A set A is a subset of a set S, written as A ⊂ S, if every element of A is also an element of S.

• If A and B are sets, then A ∩ B ⊂ A ∪ B.

• A universal set is a conveniently chosen set that contains all the elements relevant to the situation. • Let A be a subset of a universal set E. The complement of A is the set of all elements of E that are

not in A, A = {elements of E that are not in A}.




Venn diagramsA Venn diagram is a diagram used to represent the relationship between sets. For example, the four diagrams below represent four different possible relationships between two sets A and B. In each case, the universal set is again E = {1, 2, 3, …, 10}.

12 THE COUNTING RULE FOR FINITE SETS

• Let A and B be finite sets. Then

∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣

• Let A and B be disjoint finite sets. Then ∣A ∩ B∣ = 0, so:

∣A ∪ B∣ = ∣A∣ + ∣B∣

The counting rule for finite sets

To calculate the size of the union A ∪ B of two finite sets, adding the sizes of A and of B will not do, because the members of the intersection A ∩ B would be counted twice. Hence ∣A ∩ B ∣ needs to be subtracted again, and the rule is

∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣ .

For example, the Venn diagram to the right shows the sets

A = {1, 3, 4, 5, 9} and B = {2, 4, 6, 7, 8, 9}.

From the diagram, ∣A ∪ B ∣ = 9, ∣A∣ = 5, ∣B∣ = 6 and ∣A ∩ B∣ = 2, and the formula works because

9 = 5 + 6 − 2.

When two sets are disjoint, there is no overlap between A and B to cause any double counting. With ∣A ∩ B∣ = 0, the counting rule becomes

∣A ∪ B∣ = ∣A∣ + ∣B∣ .

AB

5 71

324

68

9

E

A = {1, 3, 5, 7} B = {1, 2, 3, 4}

A = {1, 3, 5, 7} B = {2, 4, 6, 8}

A = {1, 2, 3} B = {1, 2, 3, 4, 5}

A = {1, 3, 5, 7, 9} B = {1, 3}

AB5

7

123 4

6 8 9 10

E

5

7

13

2 4

8866

9 0101

EA B

AB

5

7

1

32

4

98610

E

A B5

7

13

2 4

6

8

910

E

AB

E

AB

E

AB

E

AB

E

A ∩ B A A ∩ B A ∩ B

Sets can also be visualised by shading regions of the Venn diagram, as in the following examples:




Problem solving using Venn diagramsA Venn diagram is a very convenient way to sort out problems involving overlapping sets. But once there are more than a handful of elements, listing the elements is most inconvenient. Instead, we work only with the number of elements in each region, and show that number, in brackets, in the corresponding region.

Example 9 10B

100 Sydneysiders were surveyed to find out how many of them had visited the cities of Melbourne and Brisbane. The survey showed that 31 people had visited Melbourne, 26 people had visited Brisbane and 12 people had visited both cities. Find how many people had visited:a Melbourne or Brisbane b Brisbane but not Melbournec only one of the two cities d neither city.

SOLUTION

Let M be the set of people who have visited Melbourne, let B be the set of people who have visited Brisbane, and let E be the universal set of all people surveyed. Calculations should begin with the 12 people in the intersection of the two regions. Then the numbers shown in the other three regions of the Venn diagram can easily be found.

a Number visiting Melbourne or Brisbane = 19 + 14 + 12 = 45

b Number visiting Brisbane but not Melbourne = 14

c Number visiting only one city = 19 + 14 = 33

d Number visiting neither city = 100 − 45 = 55

MB

(14)(19) (12)

(55)

E

Exercise 10C

1 Write down the following sets by listing their members.a Odd positive integers less than 10.b The first six positive multiples of 6.c The numbers on a die.d The factors of 20.

2 Find A ∪ B and A ∩ B for each pair of sets.a A = {1, 3, 5}, B = {3, 5, 7}

b A = {1, 3, 4, 8, 9}, B = {2, 4, 5, 6, 9, 10}

c A = {h, o, b, a, r, t}, B = {b, i, c, h, e, n, o}

d A = {j, a, c, k}, B = {e, m, m, a}

e A = {prime numbers less than 10}, B = {odd numbers less than 10}

3 If A = {1, 4, 7, 8} and B = {1, 2, 4, 5, 7}, state whether the following statements are true or false.a A = B b ∣A∣ = 4 c ∣B∣ = 6d A ⊂ B e A ∪ B = {1, 2, 4, 5, 7, 8} f A ∩ B = {1, 4, 7}

FOUNDATION




4 If A = {1, 3, 5}, B = {3, 4} and E = {1, 2, 3, 4, 5}, find:a ∣A∣ b ∣B∣ c A ∪ B d ∣A ∪ B∣

e A ∩ B f ∣A ∩ B∣ g A h B

5 If A = {students who study Japanese} and B = {students who study History}, carefully describe each of the following sets.a A ∩ B b A ∪ B

6 If A = {students with blue eyes} and B = {students with blond hair}, with universal set E = {students at Clarence High School}, carefully describe each of the following:a A b B c A ∪ B d A ∩ B

7 List all the subsets of each of these sets.a {a} b {a, b} c {a, b, c} d ∅

8 State in each case whether or not A ⊂ B (i.e. whether A is a subset of B).a A = {2, 3, 5}, B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

b A = {3, 6, 9, 12}, B = {3, 5, 9, 11}

c A = {d, a, n, c, e}, B = {e, d, u, c, a, t, i, o, n}

d A = {a, m, y}, B = {s, a, r, a, h}

e A = ∅ , B = {51, 52, 53, 54}

DEVELOPMENT

9 Let A = {1, 3, 7, 10} and B = {4, 6, 7, 9}, and take the universal set to be the set E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. List the members of:a A b B c A ∩ Bd A ∩ B e A ∪ B f A ∪ B

10 Let A = {1, 3, 6, 8} and B = {3, 4, 6, 7, 10}, and take the universal set to be the set E = {1, 2, 3, …, 10}. List the members of:a A b B c A ∪ B d A ∩ Be A ∪ B f A ∪ B g A ∩ B h A ∩ B

11 In each part, draw a copy of the diagram to the right and shade the corresponding region.a A ∩ B b A ∩ Bc A ∩ B d A ∪ Be A ∪ B f ∅ (∅ is the empty set)

12 Answer true or false:a If A ⊂ B and B ⊂ A, then A = B.b If A ⊂ B and B ⊂ C, then A ⊂ C.

13 Copy and complete:a If P ⊂ Q, then P ∪ Q = …b If P ⊂ Q, then P ∩ Q = …

AB

S




14 Select the Venn diagram that best shows the relationship between each pair of sets A and B.

AB

I

A B

II

AB

III

AB

IV

a A = {Tasmania}, B = {states of Australia}

b A = {7, 1, 4, 8, 3, 5}, B = {2, 9, 0, 7}

c A = {l, e, a, r, n}, B = {s, t, u, d, y}

d A = {politicians in Australia}, B = {politicians in NSW}

15 a Explain the counting rule ∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣ by making reference to the Venn diagram opposite.

b If ∣A ∪ B∣ = 17, ∣A∣ = 12 and ∣B∣ = 10, find ∣A ∩ B∣ .c Show that the relationship in part a is satisfied when

A = {3, 5, 6, 8, 9} and B = {2, 3, 5, 6, 7, 8}.

16 Use a Venn diagram to solve each problem.a In a group of 20 people, there are 8 who play the piano, 5 who play the violin and 3 who play both.

How many people play neither instrument?b Each person in a group of 30 plays either tennis or golf. 17 play tennis, while 9 play both. How many

play golf?c In a class of 28 students, there are 19 who like geometry and 16 who like trigonometry. How many

students like both if there are 5 students who don’t like either?

CHALLENGE

17 Shade each of the following regions on the given diagram (use a separate copy of the diagram for each part).a P ∩ Q ∩ Rb (P ∩ R) ∪ (Q ∩ R)

c P ∪ Q ∪ R

18 A group of 80 people was surveyed about their approaches to keeping fit. It was found that 20 jog, 22 swim and 18 go to the gym on a regular basis. Further questioning found that 10 people both jog and swim, 11 people both jog and go to the gym, and 6 people both swim and go to the gym.Finally 43 people do none of these activities.How many of the people do all three?

AB

P Q

R



10D Venn diagrams and the addition theorem 427

Venn diagrams and the addition theorem

In probability situations where an event is described using the logical words ‘and’, ‘or’ and ‘not’, Venn diagrams and the language of sets are a useful way to visualise the sample space and the event space.

A uniform sample space S is taken as the universal set because it includes all the equally likely possible outcomes. The event spaces are then subsets of S.

Mutually exclusive events and disjoint setsTwo events A and B are called mutually exclusive if they cannot both occur. For example:

If a die is thrown, the events ‘throwing a number less than three’ and ‘throwing a number greater than four’ cannot both occur and so are mutually exclusive.

If a card is drawn at random from a pack, the events ‘drawing a red card’ and ‘drawing a spade’ cannot both occur and so are mutually exclusive.

In the Venn diagram of such a situation, the two events A and B are represented as disjoint sets (disjoint means that their intersection is empty). The event ‘A and B’ is impossible, and therefore has probability zero.On the other hand, the event ‘A or B’ is represented on the Venn diagram by the union A ∪ B of the two sets. Because ∣A ∪ B∣ = ∣A∣ + ∣B∣ for disjoint sets, it follows that

P (A or B) =∣A ∪ B∣

∣S∣

=∣A∣ + ∣B∣

∣S∣ (because A and B are disjoint)

= P (A) + P (B).

10D

AB

S

13 MUTUALLY EXCLUSIVE EVENTS

Suppose that two mutually exclusive events A and B are subsets of the same uniform sample space S. Then the event ‘A or B’ is represented by A ∪ B, and

P (A or B) = P (A) + P (B).

The event ‘A and B’ cannot occur, and has probability zero.

Example 10 10D

If a die is thrown, find the probability that it is less than three or greater than four.

SOLUTION

The events ‘throwing a number less than three’ and ‘throwing a number greater than four’ are mutually exclusive,

so P (less than three or greater than four) = P (less than three) + P (greater than four) = 1

3+ 1

3

= 23

.



Chapter 10 Probability428 10D

The events ‘A and B ’ and ‘A or B ’ — the addition ruleMore generally, suppose that two events A and B are subsets of the same uniform sample space S, not necessarily mutually exclusive. The Venn diagram of the situation now represents the two events A and B as overlapping sets within the same universal set S.

The event ‘A and B’ is represented by the intersection A ∩ B of the two sets, and the event ‘A or B’ is represented by the union A ∪ B.

The general counting rule for sets is ∣A ∪ B∣ = ∣A∣ + ∣B∣ − ∣A ∩ B∣ , because the members of the intersection A ∩ B are counted in A, and are counted again in B, and so have to be subtracted. It follows then that

P (A or B) =∣A ∪ B ∣

∣S ∣

=∣A∣ + ∣B∣ − ∣A ∩ B∣

∣S∣ = P (A) + P (B) − P (A and B).

This rule is the addition rule of probability.

AB

S

Example 11 10D

If a card is drawn at random from a standard pack, find the probability that it is a red card or a spade.

SOLUTION

‘Drawing a red card’ and ‘drawing a spade’ are mutually exclusive,

so P (a red card or a spade) = P (a red card) + P (a spade) = 1

2+ 1

4 = 3

4 .

Example 12 [Mutually exclusive events in multi-stage experiments] 10D

If three coins are tossed, find the probability of throwing an odd number of tails.

SOLUTION

Let A be the event ‘one tail’ and B the event ‘three tails’. Then A and B are mutually exclusive, with

A = {HHT, HTH, THH} and B = {TTT}.

The full sample space has eight members altogether (Question 7 in the Exercise 10B lists them all), so:

P (A or B) = 38

+ 18

= 12

.

HHTHTHTHH

TTT

A B

HHHHTT

THT TTH




The symbols ∩ and ∪, and the words ‘and’, ‘or’ and ‘but’As explained in the previous section, the word ‘and’ is closely linked with the intersection of event spaces. For this reason, the event A and B is also written as A ∩ B, and the following two notations are used interchangeably:

P (A and B) means the same as P (A ∩ B) .

Similarly, the word ‘or’ is closely linked with the union of event spaces — ‘or’ always means ‘and/or’ in logic and mathematics. Thus the event A or B is also written as A ∪ B, and the following two notations are used interchangeably:

P (A or B) means the same as P (A ∪ B) .

The word ‘but’ has the same logical meaning as ‘and’ — the difference in meaning between ‘and’ and ‘but’ is rhetorical, not logical. Using this notation, the addition rule now looks much closer to the counting rule for finite sets:

P (A ∪ B) = P (A) + P (B) − P (A ∩ B).

14 THE EVENTS ‘A OR B ’ AND ‘A AND B ’

Suppose that two events A and B are subsets of the same uniform sample space S. • The event ‘A and B’ is represented by the intersection A ∩ B. • The event ‘A or B’ is represented by the union A ∪ B.

The addition ruleP (A or B) = P (A) + P (B) − P (A and B)

Example 13 10D

In a class of 30 girls, 13 play tennis and 23 play netball. If 7 girls play both sports, what is the probability that a girl chosen at random plays neither sport?

SOLUTION

Let T be the event ‘she plays tennis’, and let N be the event ‘she plays netball’.

Then P (T ) = 1330

P (N ) = 2330

and P (N ∩ T ) = 730 (meaning P (N and T )).

Hence P (N ∪ T ) = 1330

+ 2330

− 730 (meaning P (N or T ))

= 2930

,

and P (neither sport) = 1 − P (N or T)

= 130

.

Note: An alternative approach is shown in the diagram above. Starting with the 7 girls in the intersection, the numbers 6 and 16 can then be written into the respective regions ‘tennis but not netball’ and ‘netball but not tennis’. These numbers add to 29, leaving only one girl playing neither tennis nor netball.

TN

(16)

(1)

(6) (7)




Complementary events and the addition ruleIn the following example, the addition rule has to be applied in combination with the idea of complementary events. Some careful thinking is required when the words ‘and’ and ‘or’ are combined with ‘not’.

Example 14 10D

A card is drawn at random from a pack.a Find the probability that it is not an ace, but also not a two.b Find the probability that it is an even number, or a court card, or a red card (the court cards are jack,

queen and king).

Note: The word ‘or’ always means ‘and/or’ in logic and mathematics. Thus in part b, there is no need to add ‘or any two of these, or all three of these’.

SOLUTION

a The complementary event E is drawing a card that is an ace or a two.

There are eight such cards, so P (ace or two) = 852

= 213

.

Hence P (not an ace, but also not a two) = 1 − 213

= 1113

.

(The words ‘and’ and ‘but’ have different emphasis, but they have the same logical meaning.)

b The complementary event E is drawing a card that is a black odd number less than 10. This complementary event has 10 members:

E = { A♣, 3♣, 5♣, 7♣, 9♣, A♠, 3♠, 5♠, 7♠, 9♠ }.

There are 52 possible cards to choose, so:

P (E) = 1052

= 526

.

Hence, using the complementary event formula:

P (E ) = 1 − P (E) = 21

26 .




Exercise 10D

1 A die is rolled. If n denotes the number on the uppermost face, find:a P (n = 5) b P (n ≠ 5)c P (n = 4 or n = 5) d P (n = 4 and n = 5)e P (n is even or odd) f P (n is neither even nor odd)g P (n is even and divisible by three) h P (n is even or divisible by three)

2 A card is selected from a regular pack of 52 cards. Find the probability that the card:a is a jack b is a 10c is a jack or a 10 d is a jack and a 10e is neither a jack nor a 10 f is blackg is a picture card h is a black picture cardi is black or a picture card j is neither black nor a picture card.

3 Show that the following events A and B are disjoint by showing that A ∩ B = ∅, and check that P (A or B) = P (A) + P (B).a Two coins are thrown and A = {two heads}, B = {one tail}.b A committee of two is formed by choosing at random from two men, Ricardo and Steve, and one

woman, Tania. Suppose A = {committee of two men}, B = {committee includes Tania}.

4 A die is thrown. Let A be the event that an even number appears. Let B be the event that a number greater than 2 appears.a Are A and B mutually exclusive?b Find:

i P (A) ii P (B) iii P (A ∩ B) iv P (A ∪ B)

5 Two dice are thrown. Let a and b denote the numbers rolled. Find:a P (a is odd) b P (b is odd) c P (a and b are odd)d P (a or b is odd) e P (neither a nor b is odd ) f P (a = 1)g P (b = a) h P (a = 1 and b = a) i P (a = 1 or b = a)j P (a ≠ 1 and a ≠ b)

FOUNDATION




DEVELOPMENT

6 a A

B

P (A) = 12 and P (B) = 1

3.

Find:i P (A)ii P (B)iii P (A and B) iv P (A or B) v P (neither A nor B)

b A B

P (A) = 25 and P (B) = 1

5.

Find:i P (A)ii P (B)iii P (A or B)iv P (A and B) v P (not both A and B)

c A B

P (A) = 12 , P (B) = 1

3 and

P (A and B) = 16.

Find:

i P (A)ii P (B)iii P (A or B)iv P (neither A nor B)v P (not both A and B)

7 Use the addition rule P (A ∪ B) = P (A) + P (B) − P (A ∩ B) to answer the following questions.

a If P (A) = 15 , P (B) = 1

3 and P (A ∩ B) = 1

15 , find P (A ∪ B).

b If P (A) = 12 , P (B) = 1

3 and P (A ∪ B) = 5

6 , find P (A ∩ B).

c If P (A ∪ B) = 910

, P (A ∩ B) = 15 and P (A) = 1

2 , find P (B).

d If A and B are mutually exclusive and P (A) = 17 and P (B) = 4

7 , find P (A ∪ B).

8 An integer n is picked at random, where 1 ≤ n ≤ 20. The events A, B, C and D are:A: an even number is chosen,B: a number greater than 15 is chosen,C: a multiple of 3 is chosen,D: a one-digit number is chosen.

a i Are the events A and B mutually exclusive?ii Find P (A), P (B) and P (A and B) and hence evaluate P (A or B).

b i Are the events A and C mutually exclusive?ii Find P (A), P (C) and P (A and C) and hence evaluate P (A or C).

c i Are the events B and D mutually exclusive?ii Find P (B), P (D) and P (B and D), and hence evaluate P (B or D).

9 In a group of 50 students, there are 26 who study Latin and 15 who study Greek and 8 who study both languages. Draw a Venn diagram and find the probability that a student chosen at random:a studies only Latinb studies only Greekc does not study either language.




10 During a game, all 21 members of an Australian Rules football team consume liquid. Some players drink only water, some players drink only GatoradeTM and some players drink both. There are 14 players who drink water and 17 players who drink GatoradeTM.a How many drink both water and GatoradeTM?b If one team member is selected at random, find the probability that:

i he drinks water but not GatoradeTM,ii he drinks GatoradeTM but not water.

11 Each student in a music class of 28 studies either the piano or the violin or both. It is known that 20 study the piano and 15 study the violin. Find the probability that a student selected at random studies both instruments.

CHALLENGE

12 List the 25 primes less than 100. A number is drawn at random from the integers from 1 to 100. Find the probability that:a it is primeb it has remainder 1 after division by 4c it is prime and it has remainder 1 after division by 4d it is either prime or it has remainder 1 after division by 4.

13 A group of 60 students was invited to try out for three sports: rugby, soccer and cross country. Of these, 32 tried out for rugby, 29 tried out for soccer, 15 tried out for cross country, 11 tried out for rugby and soccer, 9 tried out for soccer and cross country, 8 tried out for rugby and cross country, and 5 tried out for all three sports. Draw a Venn diagram and find the probability that a student chosen at random:a tried out for only one sportb tried out for exactly two sportsc tried out for at least two sportsd did not try out for a sport.



434 Chapter 10 Probability 10E

Multi-stage experiments and the product rule

This section deals with experiments that have a number of stages. The full sample space of such an experiment can quickly become too large to be conveniently listed, but instead a rule can be developed for multiplying together the probabilities associated with each stage.

Two-stage experiments — the product ruleHere is a simple question about a two-stage experiment:

‘Throw a die, then toss a coin. What is the probability of obtaining at least 2 on the die, followed by a head?’

Graphed to the right are the twelve possible outcomes of the experiment, all equally likely, with a box drawn around the five favourable outcomes. Thus

P (at least 2 and a head) = 512

.

Now consider the two stages separately. The first stage is throwing a die, and the desired outcome is A = ‘getting at least 2’ — here there are six possible outcomes and five favourable outcomes, giving

probability 56. The second stage is tossing a coin, and the desired outcome is B = ‘tossing a head’ — here

there are two possible outcomes and one favourable outcome, giving probability 12.

The full experiment then has 6 × 2 = 12 possible outcomes and there are 5 × 1 = 5 favourable outcomes. Hence

P (AB) = 5 × 16 × 2

= 56

× 12

= P (A) × P (B).

Thus the probability of the compound event ‘getting at least 2 and a head’ can be found by multiplying together the probabilities of the two stages. The argument here can easily be generalised to any two-stage experiment.

Independent eventsThe word ‘independent’ needs further discussion. In the example above, the throwing of the die clearly does not affect the tossing of the coin, so the two events are independent.

here is a very common and important type of experiment where the two stages are not independent:

‘Choose an elector at random from the NSW population. First note the elector’s gender. Then ask the elector if he or she voted Labor or non-Labor in the last State election.’

In this example, one might suspect that the gender and the political opinion of a person may not be independent and that there is correlation between them. This is in fact the case, as almost every opinion poll has shown over the years. More on this in Year 12.

10E

H

654321

T

die

coin

15 TWO-STAGE EXPERIMENTS

If A and B are events in successive independent stages of a two-stage experiment, thenP (AB) = P (A) × P (B)

where the word ‘independent’ means that the outcome of one stage does not affect the probabilities of the other stage.



10E Multi-stage experiments and the product rule 435

Multi-stage experiments — the product ruleThe same arguments clearly apply to an experiment with any number of stages.

Example 15 10E

A pair of dice is thrown twice. What is the probability that the first throw is a double, and the second throw gives a sum of at least 4?

SOLUTION

We saw in Section 10B that when two dice are thrown, there are 36 possible outcomes, graphed in the diagram to the right.There are six doubles among the 36 possible outcomes,

so P (double) = 636

= 1

6 .

All but the pairs (1, 1), (2, 1) and (1, 2) give a sum at least 4,

so P (sum is at least 4) = 3336

= 11

12 .

Because the two stages are independent,

P (double, sum at least 4) = 16

× 1112

= 11

72 .

1

3

5

2

4

6

1stthrow

2nd throw

654321

15 MULTI-STAGE EXPERIMENTS

If A1, A2, …, An are events in successive independent stages of a multi-stage experiment, then

P (A1 A2 …

An) = P (A1) × P (A2) × … × P (An).

Example 16 10E

A coin is tossed five times. Find the probability that:a every toss is a headb no toss is a headc there is at least one head.




Sampling without replacement — an extension of the product ruleThe product rule can be extended to the following question, where the two stages of the experiment are not independent.

SOLUTION

The five tosses are independent events.

a P (HHHHH) = 12

× 12

× 12

× 12

× 12

= 1

32

b P (no heads) = P (T T T T T) = 1

2× 1

2× 1

2× 1

2× 1

2 = 1

32

c P (at least one head) = 1 − P (no heads) = 1 − 1

32 = 31

32

Example 17 10E

A box contains five discs numbered 1, 2, 3, 4 and 5.Two numbers are drawn in succession, without replacement.What is the probability that both are even?

SOLUTION

The probability that the first number is even is 25 .

When this even number is removed, one even and three odd numbers

remain, so the probability that the second number is also even is 14 .

Hence P (both even) = 25

× 14

= 1

10 .

Note: The graph to the right allows the calculation to be checked by examining its full sample space.

Because doubles are not allowed (i.e. there is no replacement), there are only 20 possible outcomes.

The two boxed outcomes are the only outcomes that consist of two even numbers, giving the same

probability of 220

= 110

.

1

1

3

5

2

4

1st

number

2nd number

2 3 4 5




Listing the favourable outcomesThe product rule is often combined with a listing of the favourable outcomes. A tree diagram may help in producing that listing, although this is hardly necessary in the straightforward example below, which continues an earlier example.

Example 18 10E

A coin is tossed four times. Find the probability that:a the first three coins are heads b the middle two coins are tails.

c Copy and complete this table:

SOLUTION

a P (the first three coins are heads) = P (HHHH) + P (HHHT)

(notice that the two events HHHH and HHHT are mutually exclusive)

= 116

+ 116

(because each of these two probabilities is 12

× 12

× 12

× 12)

= 18

.

b P (middle two are tails) = P (HTTH) + P (HTTT) + P (TTTH) + P (TTTT)

= 116

+ 116

+ 116

+ 116

= 14

.

c P (no heads) = P (TTTT)

= 116

P (one head) = P (HTTT) + P (THTT) + P (TTHT) + P (TTTH)

= 416

P (two heads) = P (HHTT) + P (HTHT) + P (THHT) + P (HTTH) + P (THTH) + P (TTHH)

(these are all six possible orderings of H, H, T and T)

= 616

, and the last two results follow by symmetry.

Number of heads 0 1 2 3 4

Probability116

416

616

416

116

Number of heads 0 1 2 3 4

Probability




Exercise 10E

1 Suppose that A, B, C and D are independent events, with P (A) = 18 , P (B) = 1

3 , P (C) = 1

4

and P (D) = 27 . Use the product rule to find:

a P (AB) b P (AD) c P (BC) d P (ABC) e P (BCD) f P (ABCD)

2 A coin and a die are tossed. Use the product rule to find the probability of obtaining:a a three and a head b a six and a tailc an even number and a tail d a number less than five and a head

3 One set of cards contains the numbers 1, 2, 3, 4 and 5, and another set contains the letters A, B, C, D and E. One card is drawn at random from each set. Use the product rule to find the probability of drawing:a 4 and B b 2 or 5, then Dc 1, then A or B or C d an odd number and Ce an even number and a vowel f a number less than 3, and Eg the number 4, followed by a letter from the word MATHS.

4 Two marbles are picked at random, one from a bag containing three red and four blue marbles, and the other from a bag containing five red and two blue marbles. Find the probability of drawing:a two red marbles b two blue marblesc a red marble from the first bag and a blue marble from the second.

5 A box contains five light bulbs, two of which are faulty. Two bulbs are selected, one at a time without replacement. Find the probability that:a both bulbs are faulty b neither bulb is faultyc the first bulb is faulty and the second one is not d the second bulb is faulty and the first one is not.

FOUNDATION

Example 19 10E

Wes is sending Christmas cards to ten friends. He has two cards with angels, two with snow, two with reindeer, two with trumpets and two with Santa Claus.What is the probability that Harry and Helmut get matching cards?

SOLUTION

Describe the process in a different way as follows:

‘Wes decides that he will choose Harry’s card first and Helmut’s card second. Then he will choose the cards for his remaining eight friends.’

All that matters now is whether the card that Wes chooses for Helmut is the same as the card that he has already chosen for Harry. After he chooses Harry’s card, there are nine cards remaining, of which only one

matches Harry’s card. Thus the probability that Helmut’s card matches is 19 .

Describing the experiment in a different waySometimes, the manner in which an experiment is described makes calculation difficult, but the experiment can be described in a different way so that the probabilities are the same but the calculations are much simpler.




DEVELOPMENT

6 A die is rolled twice. Using the product rule, find the probability of throwing:a a six and then a five b a one and then an odd numberc a double six d two numbers greater than foure a number greater than four and then a number less than four.

7 A box contains 12 red and 10 green discs. Three discs are selected, one at a time without replacement.a What is the probability that the discs selected are red, green, red in that order?b What is the probability of this event if the disc is replaced after each draw?

8 a From a standard pack of 52 cards, two cards are drawn at random without replacement. Find the probability of drawing:i a spade and then a heartiii a jack and then a queen

ii two clubsiv the king of diamonds and then the ace of clubs.

b Repeat the question if the first card is replaced before the second card is drawn.

9 A coin is weighted so that it is twice as likely to fall heads as it is tails.a Write down the probabilities that the coin falls:

i heads ii tails.b If you toss the coin three times, find the probability of:

i three heads ii three tails iii head, tail, head in that order.

10 If a coin is tossed repeatedly, find the probability of obtaining at least one head in:a two tosses b five tosses c ten tosses.

11 [Valid and invalid arguments]Identify any fallacies in the following arguments. If possible, give some indication of how to correct them.a ‘The probability that a Year 12 student chosen at random likes classical music is 50%, and the

probability that a student plays a classical instrument is 20%. Therefore the probability that a student chosen at random likes classical music and plays a classical instrument is 10%.’

b ‘The probability of a die showing a prime is 12 , and the probability that it shows an odd number is 1

2 .

Hence the probability that it shows an odd prime number is 12

× 12

= 14 .’

c ‘I choose a team at random from an eight-team competition. The probability that the team wins any game is 1

2 , so the probability that it defeats all of the other seven teams is (

12)

7= 1

128 .’

d ‘A normal coin is tossed and shows heads eight times. Nevertheless, the probability that it shows heads the next time is still 1

2 .’

12 A die is rolled twice. Using the product rule, find the probability of rolling:a a double two b any doublec a number greater than three, then an odd

numberd a one and then a four

e a four and then a one f a one and a four in any orderg an even number, then a five h a five and then an even numberi an even number and a five in any order.




13 An archer fires three shots at a bullseye. He has a 90% chance of hitting the bullseye. Using H for hit and M for miss, list all eight possible outcomes. Then, assuming that successive shots are independent, use the product rule to find the probability that he will:a hit the bullseye three times b miss the bullseye three timesc hit the bullseye on the first shot only d hit the bullseye exactly oncee miss the bullseye on the first shot only f miss the bullseye exactly once.(Hint: Part d requires adding the probabilities of HMM, MHM and MMH, and part f requires a similar calculation.)

CHALLENGE

14 One layer of tinting material on a window cuts out 15 of the sun’s UV light coming through it.

a What fraction would be cut out by using two layers?b How many layers would be required to cut out at least 9

10 of the UV light?

c What relation does this question have to questions about probability?

15 In a lottery, the probability of the jackpot prize being won in any draw is 160

.

a What is the probability that the jackpot prize will be won in each of four consecutive draws?b How many consecutive draws need to be made for there to be a greater than 98% chance that at least

one jackpot prize will have been won?

16 (This question is best done by retelling the story of the experiment, as explained in the notes above.) Nick has five different pairs of socks to last the working week, and they are scattered loose in his drawer. Each morning, he gets up before light and chooses two socks at random. Find the probability that he wears a matching pair:a on the first morning b on the last morningc on the third morning d the first two morningse every morning f every morning but one.



441

Probability tree diagrams

In more complicated problems, and particularly in unsymmetric situations, a probability tree diagram can be very useful in organising the various cases, in preparation for the application of the product rule.

Constructing a probability tree diagramA probability tree diagram differs from the tree diagrams used in Section 10B for counting possible outcomes, in that the relevant probabilities are written on the branches and then multiplied together in accordance with the product rule. An example will demonstrate the method.

As before, these diagrams have one column labelled ‘Start’, a column for each stage, and a column listing the outcomes, but there is now an extra column labelled ‘Probability’ at the end.

10F

Example 20 10F

One card is drawn from each of two packs. Use a probability tree diagram to find the probability that:a both cards are court cards ( jack, queen or king)b neither card is a court cardc one card is a court card and the other is not.

SOLUTION

In each pack of cards, there are 12 court cards out of 52 cards,

so P (court card) = 313

and P (not a court card) = 1013

.

1stDraw

2ndDraw Outcome Probability

Start

CC

CC

C

C

CC

CC

CC

CC

91693016930169

100169

313

1013

313

1013

313

1013

Multiply the probabilities along each arm because they are successive stages.Add the probabilities in the final column because they are mutually exclusive.

a P (two court cards) = 313

× 313

= 9169

b P (no court cards) = 1013

× 1013

= 100169

c P (one court card) = 313

× 1013

+ 1013

× 313

= 30169

+ 30169

= 60169

Note: The four probabilities in the last column of the tree diagram add exactly to 1, which is a useful check on the working. The three answers also add to 1.

10F Probability tree diagrams



442 10FChapter 10 Probability

Example 21 [A more complicated experiment] 10F

A bag contains six white marbles and four blue marbles. Three marbles are drawn in succession. At each draw, if the marble is white it is replaced, and if it is blue it is not replaced.

Find the probabilities of drawing:a no blue marbles b one blue marblec two blue marbles d three blue marbles.

SOLUTION

With the ten marbles all in the bag:

P (W) = 610

= 35 and P (B) = 4

10= 2

5.

If one blue marble is removed, there are six white and three blue marbles, so

P (W) = 69

= 23 and P (B) = 3

9= 1

3 .

If two blue marbles are removed, there are six white and two blue marbles, so

P (W) = 68

= 34 and P (B) = 2

8= 1

4 .

Start1st

draw2nddraw

3rddraw

W

B

W

B

W

B

W

B

W

B

W

B

W

B

WWW

WWB

WBW

WBB

BWW

BWB

BBW

BBB

35

13

14

25

23

34

13

23

13

23

25

35

25

35

27125

18125

425

225

845

445

110

130

Outcome Probability

In each part, multiply the probabilities along each arm and then add the cases.

a P (no blue marbles) = 27125

b P (one blue marble) = 18125

+ 425

+ 845

= 5421125

c P (two blue marbles) = 225

+ 445

+ 110

= 121450

d P (three blue marbles) = 130

Note: Again, as a check on the working, your calculator will show that the eight probabilities in the last column of the diagram add exactly to 1 and that the four answers above also add to 1.

17 PROBABILITY TREE DIAGRAMS

In a probability tree diagram: • Probabilities are written on the branches. • There is a final column giving the probability of each outcome.



443

Exercise 10F

1 A bag contains three black and four white discs. A disc is selected from the bag, its colour is noted, and the disc is then returned to the bag before a second disc is drawn.a By multiplying along the branches of the tree, find:

i P (BB) ii P (BW)iii P (WB) iv P (WW)

b Hence, by adding, find the probability that:i both discs drawn have the same colourii the discs drawn have different colours.

c Draw your own tree diagram, and repeat part b if the first disc is not replaced before the second one is drawn.

2 Two light bulbs are selected at random from a large batch of bulbs in which 5% are defective.a By multiplying along the branches of the tree, find:

i P (both bulbs work)

ii P (the first works, but the second is defective)iii P (the first is defective, but the second works)iv P (both bulbs are defective).

b Hence find the probability that at least one bulb works.

3 One bag contains three red and two blue balls and another bag contains two red and three blue balls. A ball is drawn at random from each bag.a By multiplying along the branches of the tree, find:

i P (RR) ii P (RB)iii P (BR) iv P (BB)

b Hence, by adding, find the probability that:i the balls are the same colourii the balls are different colours.

4 In group A there are three girls and seven boys, and in group B there are six girls and four boys. One person is chosen at random from each group. Draw a probability tree diagram.a By multiplying along the branches of the tree, find:

i P (GG) ii P (GB) iii P (BG) iv P (BB)b Hence, by adding, find the probability that:

i two boys or two girls are chosen ii one boy and one girl are chosen.

5 There is an 80% chance that Garry will pass his driving test and a 90% chance that Emma will pass hers. Draw a probability tree diagram, and find the chance that:a Garry passes and Emma failsb Garry fails and Emma passesc only one of Garry and Emma passesd at least one fails.

FOUNDATION

Start

W

B

W

B

W

B

1stDraw

2ndDraw

37

37

37

47

47

47

Start

W

D

W

D

W

D

1stBulb

2ndBulb

0.05

0.05

0.050.95

0.95

0.95

Start

R

B

R

B

R

B

Bag 1

25

25

25

35

35

35

Bag 2




444 10FChapter 10 Probability

DEVELOPMENT

6 The probability that a set of traffic lights will be green when you arrive at them is 35 . A motorist drives

through two sets of lights. Assuming that the two sets of traffic lights are not synchronised, find the probability that:a both sets of lights will be greenb at least one set of lights will be green.

7 A factory assembles calculators. Each calculator requires a chip and a battery. It is known that 1% of chips and 4% of batteries are defective. Find the probability that a calculator selected at random will have at least one defective component.

8 The probability of a woman being alive at 80 years of age is 0.2, and the probability of her husband being alive at 80 years of age is 0.05. Find the probability that:a they will both live to be 80 years of ageb only one of them will live to be 80.

9 Alex and Julia are playing in a tennis tournament. They will play each other twice, and each has an equal chance of winning the first game. If Alex wins the first game, his probability of winning the second game is increased to 0.55. If he loses the first game, his probability of winning the second game is reduced to 0.25. Find the probability that Alex wins exactly one game.

10 One bag contains four red and three blue discs, and another bag contains two red and five blue discs. A bag is chosen at random and then a disc is drawn from it. Find the probability that the disc is blue.

11 In a raffle in which there are 200 tickets, the first prize is drawn and then the second prize is drawn without replacing the winning ticket. If you buy 15 tickets, find the probability that you win:a both prizes b at least one prize.

12 A box contains 10 chocolates, all of identical appearance. Three of the chocolates have caramel centres and the other seven have mint centres. Hugo randomly selects and eats three chocolates from the box. Find the probability that:a the first chocolate Hugo eats is caramelb Hugo eats three mint chocolatesc Hugo eats exactly one caramel chocolate.

13 In an aviary there are four canaries, five cockatoos and three budgerigars. If two birds are selected at random, find the probability that:a both are canaries b neither is a canaryc one is a canary and one is a cockatoo d at least one is a canary.

14 Max and Jack each throw a die. Find the probability that:a they do not throw the same numberb the number thrown by Max is greater than the number thrown by Jackc the numbers they throw differ by three.

15 In a large co-educational school, the population is 47% female and 53% male. Two students are selected from the school population at random. Find, correct to two decimal places, the probability that:a both are male b a girl and a boy are selected.



445

16 The numbers 1, 2, 3, 4 and 5 are each written on a card. The cards are shuffled and one card is drawn at random. The number is noted and the card is then returned to the pack. A second card is selected, and in this way a two-digit number is recorded. For example, a 2 on the first draw and a 3 on the second results in the number 23.a What is the probability of the number 35 being recorded?b What is the probability of an odd number being recorded?

17 A twenty-sided die has the numbers from 1 to 20 on its faces.a If it is rolled twice, what is the probability that the same number appears on the uppermost face each time?b If it is rolled three times, what is the probability that the number 15 appears on the uppermost face

exactly twice?

18 An interviewer conducts a poll in Sydney and Melbourne on the popularity of the prime minister. In Sydney, 52% of the population approve of the prime minister, and in Melbourne her approval rating is 60%. If one of the two capital cities is selected at random and two electors are surveyed, find the probability that:a both electors approve of the prime ministerb at least one elector approves of the prime minister.

19 In a bag there are four green, three blue and five red discs.a Two discs are drawn at random, and the first disc is not replaced before the second disc is drawn.

Find the probability of drawing:i two red discs ii one red and one blue disciii at least one green disc iv a blue disc on the first drawv two discs of the same colour vi two differently coloured discs.

b Repeat part a if the first disc is replaced before the second disc is drawn.

CHALLENGE

20 In a game, two dice are rolled and the score given is the higher of the two numbers on the uppermost faces. For example, if the dice show a three and a five, the score is a five.a Find the probability that you score a one in a single throw of the two dice.b What is the probability of scoring three consecutive ones in three rolls of the dice?c Find the probability that you score a six in a single roll of the dice.

21 In each game of Sic Bo, three regular six-sided dice are thrown once.a In a single game, what is the probability that all three dice show six?b What is the probability that exactly two of the dice show six?c What is the probability that exactly two of the dice show the same number?d What is the probability of rolling three different numbers on the dice?

22 A bag contains two green and two blue marbles. Marbles are drawn at random, one by one without replacement, until two green marbles have been drawn. What is the probability that exactly three draws will be required?




446 10GChapter 10 Probability

Conditional probability

Sometimes our knowledge of the results of an experiment change over time, and as we gain more information, our probabilities change. Conditional probability allows us to calculate these changing probabilities.

Reduced sample space and reduced event spaceDrago will win a certain game if an odd number is thrown on a die. The sample space is S = {1, 2, 3, 4, 5, 6}, and the event space is A = {1, 3, 5}, so

P (he wins) = 36

= 12

.

Drago nervously turns away as the die is thrown, and someone calls out, ‘It’s a prime number!’ What, for him, is now the probability that he wins the game?• First, the sample space has now been reduced to the set B = {2, 3, 5} of prime numbers on the die. This

is called the reduced sample space.• Secondly, the event space has now been reduced to the set {3, 5} of odd prime numbers on the die. This

reduced event space is the intersection A ∩ B of the event space A and the reduced event space B. Hence

P (he wins, given that the number is prime) =∣ reduced event space ∣

∣ reduced sample space ∣ = 2

3 .

Alternatively, with B as the reduced sample space and A ∩ B as the reduced event space, we can write the working using a concise formula,

P (A ∣B) =∣A ∩ B∣

∣B∣ = 2

3 .

where P (A ∣B) is the symbol for conditional probability, and means ‘the probability of A, given that B has occurred’.

(Of course we are assuming that the person calling out was telling the truth.)

Conditional probabilityIt is only a matter of convenience whether we work with the reduced sample space and reduced event space, or with formulae.

10G



10G Conditional probability 447

18 CONDITIONAL PROBABILITY

Suppose that two events A and B are subsets of same uniform sample space S. Then the conditional probability P (A ∣B) of A, given that B has occurred, is obtained by removing all the elements not in B from the sample space S, and from the event space A. Thus there are two equivalent formulae,

P (A ∣B) =∣ reduced event space ∣

∣ reduced sample space∣ and P (A ∣B) = ∣A ∩ B ∣

∣B ∣.

Read P (A ∣B) as ‘the probability of A, given that B has occurred’.

Example 22 10G

A card is drawn at random from a normal pack of 52 cards. Find the probability that it is a 9, 10, jack or queen if:a Nothing more is known.b It is known to be a court card (jack, queen or king).c It is known to be an 8 or a 10.d It is known not to be a 2, 3, 4 or 5.

SOLUTION

Let the sample space S be the set of all 52 cards, and letA = {9♣, 9♦, 9♥, 9♠, 10♣, 10♦, 10♥, 10♠, J♣, J♦, J♥, J♠, Q♣, Q♦, Q♥, Q♠}.

a There is no reduction of the sample space or the event space,

so P (A) =∣A∣∣S∣

= 1652

= 413

.

b The reduced sample space isB = {court cards},

and the reduced event space is A ∩ B = {jacks and queens},

so P (A ∣B) =∣A ∩B∣

∣B∣ = 8

12

= 23

.

c The reduced sample space isB = {8s and 10s}

and the reduced event space is A ∩ B = {10s},

so P (A ∣B) =∣A ∩ B ∣

∣B ∣ = 4

8

= 12

.

d The reduced sample space isB = {cards not 2, 3, 4 or 5}

and the reduced event space is A ∩ B = A,

so P (A ∣B) =∣A ∩ B ∣

∣B ∣ = 16

36

= 49

.




Another formula for conditional probabilitySuppose as before that two events A and B are subsets of the same uniform sample space S. The previous formula gave P (A ∣B) in terms of the sizes of the sets A ∩ B and B. By dividing through by ∣S ∣ , we obtain a formula for P (A ∣B) in terms of the probabilities of A ∩ B and B:

P (A ∣B) =∣A ∩ B ∣

∣B ∣

=∣A ∩ B ∣

∣S ∣ ÷

∣B ∣∣S ∣

=P (A ∩ B)

P (B)

19 ANOTHER FORMULA FOR CONDITIONAL PROBABILITY

For two events A and B,

P (A ∣B) = P (A ∩ B)P (B)

.

The great advantage of this formula is that it can be used when nothing is known about the actual sizes of the sample space and the event space, as in the example below.

Example 23 10G

In a certain population, 35% of people have blue eyes, 15% have blond hair, and 10% have blue eyes and blond hair. A person is chosen from this population at random.a Find the probability that they have blond hair, given that they have blue eyes.b Find the probability that they have blue eyes, given that they have blond hair.

SOLUTION

From the given percentages, we know thatP (blue eyes) = 0.35, P (blond hair) = 0.15, P (blue eyes and blond hair) = 0.1 .

a P (blond hair, given blue eyes) =P (blond hair and blue eyes)

P (blue eyes)

= 0.10.35

≑ 0.29

b P (blue eyes, given blond hair) =P (blond hair and blue eyes)

P (blond hair)

= 0.10.15

≑ 0.67




Independent eventsWe say that an event A is independent of an event B if knowing whether or not B has occurred does not affect our probability of A. That is,

A and B are independent means that P (A ∣B) = P (A) .

20 INDEPENDENT EVENTS

• Two events A and B are called independent if the probability of A is not affected by knowing whether or not B has occurred:

P (A ∣B) = P (A) • If A is independent of B, then B is independent of A.

The second dot point may seem trivial, but it needs proving.Suppose that A is independent of B, meaning that P (A ∣B) = P (A) .Then to prove that B is independent of A,

P (B ∣A) =P (B ∩ A)

P (A) (using the formula for P (B ∣A))

=P (A ∩ B)

P (B)×

P (B)P (A)

(multiply top and bottom by P (B))

= P (A ∣B) ×P (B)P (A)

(using the formula for P (A ∣B))

= P (A) ×P (B)P (A)

(we are assuming that P (A ∣B) = P (A))

= P (B), as required.

The consequence of this is that we can just say that ‘the two events A and B are independent’, which is what we would expect.

Example 24 10G

Two dice are thrown one after the other. • Let A be the event ‘the first die is odd’. • Let B be the event ‘the second die is 1, 2 or 3’. • Let C be the event ‘the sum is five’.

Which of the three pairs of events are independent?

SOLUTION

The sample space S has size 36, and

P (A) = 1836

= 12

, P (B) = 1836

= 12

, P (C) = 436

= 19

.

The last result follows because 5 = 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1 are all the ways that a sum of five can be obtained. Taking intersections,

P (A ∩ B) = 936

= 14

, P (B ∩ C) = 336

= 112

, P (C ∩ A) = 236

= 118

.

1

1 2 3 4 5 6

3

5

2

4

6

1stthrow

2nd throw




Testing the definition of independence on the three pairs of events:

P (A ∣B) =P (A ∩ B)

P (B)

= 14

÷ 12

= 12

= P (A)

P (B ∣C) =P (B ∩ C)

P (C)

= 112

÷ 19

= 34

≠ P (B)

P (C ∣ A) =P (C ∩ A)

P (A)

= 118

÷ 12

= 19

= P (C)

Hence A and B are independent, and C and A are independent, but B and C are not independent.

Note: The word ‘independent’ here means ‘mathematically independent’, and carries no scientific implication. A scientist who makes a statement about the natural world, on the basis of such mathematical arguments, is doing science, not mathematics.

Note: The word ‘independent’ was introduced in Section 10E with multi-stage experiments. Its use in this section is a generalisation of its use in the multi-stage situation. For example, the events A and B above are independent in both senses, but the independence of C and A is a new idea.

The product rule

The formula P (A ∣B) = P (A ∩ B)P (B)

can be rearranged as

P (A ∩ B) = P (A ∣B) × P (B).

If the events are independent, then P (A ∣B) = P (A) , so

P (A ∩ B) = P (A) × P (B).

Conversely, if P = (A ∩ B) = P (A) × P (B) , then P (A ∣B) = P (A) , meaning that the events are independent. Thus we have both a formula that applies to independent events, and another test for independence:

21 INDEPENDENT EVENTS AND THE PRODUCT RULE

• For any events A and B, P (A and B) = P (A∣B) × P (B). • If A and B are independent events, then

P (A and B) = P (A) × P (B). • Conversely, if P (A and B) = P (A) × P (B) , then the events are independent.

Remember that P (A and B) and P (A ∩ B) mean exactly the same thing.

All this is a generalisation of the product rule introduced with the multi-stage experiments and tree diagrams in Section 10E.




Example 25 10G

Calculate P (A) × P (B) and P (B) × P (C) and P (C) × P (A) in the previous example. Then use the third dotpoint of Box 21 above to confirm that these calculations give the same results for the independence of A and B, of B and C, and of C and A.

SOLUTION

P (A) × P (B) = 12

× 12

= 14

= P (A ∩ B)

P (B) × P (C) = 12

× 19

= 118

≠ P (B ∩ C)

P (C) × P (A) = 19

× 12

= 118

= P (C ∩ A)

again showing that A and B are independent, that C and A are independent, but that B and C are not independent.

Example 26 10G

Example 23 described the incidence of blue eyes and blond hair in a population:a Why does the product rule formula show immediately that ‘blue eyes’ and ‘blond hair’ are not

independent?b What would be the probability of having blue eyes and blond hair if the two characteristics were

independent?

SOLUTION

a Box 21 says that ‘blue eyes’ and blond hair are independent if and only ifP (blue eyes and blond hair) = P (blue eyes) × P (blond hair).

But according to the data given in Example 23, LHS = 0.1 and RHS = 0.35 × 0.15 = 0.0525, so the events are not independent.

b If the events were independent, then the product rule formula would hold, so

P (blue eyes and blond hair) = P (blue eyes) × P (blond hair)

= 0.35 × 0.15 = 0.0525.

Using the sample space without any formulaeSometimes, however, it is easier not to use any of the machinery in this section, but to go back to basics and deal directly with the sample space, with little or no notation.

Example 27 10G

Yanick and Oskar can’t agree on which movie to watch, so they play a game throwing two dice. • If the sum is 7 or 8, Yanick chooses the movie. • If the sum is 9, 10, 11 or 12, Oskar chooses the movie. • If the sum is less than 7, they throw the dice again.

Who has the higher probability of choosing the movie?




SOLUTION

There are 36 possible throws in the original sample space.But we can ignore the throws whose sum is less than 7, represented by the open circles, because these 15 throws are discarded if they occur.This leaves a sample space with only the 21 closed circles.There are 6 + 5 = 11 throws with sum 7 or 8, so

P (Yanick chooses) = 1121

and P (Oskar chooses) = 1021

.

Hence Yanick has the higher probability of choosing.

1

1 2 3 4 5 6

3

5

2

4

6

1stthrow

2nd throw

A final note about general sample spacesThe theory developed in this chapter can be summarised in four results:• P (not A) = 1 − P (A)• P (A or B) = P (A) + P (B) − P (A and B)

• P (A ∣B) = P (A ∩ B)P (B)

• P (AB) = P (A) × P (B) if and only if A and B are independent events.

These four results were all developed using equally likely possible outcomes, which restricts them to experiments in which there is a finite sample space that is uniform. The four results, however, hold in all theories of probability, either as axioms and definitions, or as results proven by different methods.

Exercise 10GNote: Several questions in this exercise, and elsewhere, require the assumptions that a baby is equally likely to be a boy or a girl, that these are the only two possibilities, and that the sex of a child is independent of the sex of any previous children. These assumptions are simplifications of complex scientific and social considerations, and they do not describe the real situation.

1 Two dice are thrown. The sample space of this experiment is shown in the dot diagram alongside. Following the throw, it is revealed that the first die shows an even number.a Copy the diagram and circle the reduced sample space.b Find the conditional probability of getting two sixes.c Find the conditional probability of getting at least one six.d Find the conditional probability that the sum of the two numbers is five.

FOUNDATION

1

1 2 3 4 5 6

3

5

2

4

6

1stthrow

2nd throw




2 A poll is taken amongst 1000 people to determine their voting patterns in the last election.

Coalition Labor Other Total

Male 130 340 110 580

Female 210 190 20 420

Total 340 530 130 1000

a Determine the probability that a particular person in the group voted Coalition.b What is the probability that a particular person voted Labor, given that they were female?c What is the probability that a particular member of the group was male, if it is known that they voted

Coalition?d Robin voted neither Coalition nor Labor. What is the probability that Robin was female?

3 A student is investigating if there is any relationship between those who choose Mathematics Extension 1 (M) and those who choose English Extension 1(E) at his school.

a Copy and complete the table by filling in the totals.b Find the probability that a particular student chose:

i neither Maths Extension 1 nor English Extension 1?ii English Extension 1, given that they chose Maths Extension 1?iii Maths Extension 1, if it is known that they chose English Extension 1?iv Maths Extension 1, given that they did not choose English Extension 1?

4 Two cards are drawn from a standard pack — the first card is replaced and the pack shuffled before the second card is drawn — and the suit of each card (S, H, D or C) is noted by the game master. A player wants to know the probability that both cards are hearts.a What is the probability if nothing else is known?b The first card is known to be a heart. List the reduced sample space. What is the conditional

probability of two hearts?c List the reduced sample space if at least one of the cards is known to be a heart. What is the

probability of two hearts?d List the reduced sample space if the first card is known to be red. What is the probability of two hearts?

5 In a certain game, the player tosses two coins and then throws a die. The sample space is shown in the table.

The rules of the game assign one point for each head, and zero points for each tail. This score is then added to the score on the die.a Copy the table and fill in the point score for each outcome.b Find the probability that a player gets more than 7 points.c Suppose it is known that he has thrown two heads. Find the probability that a player gets more than

7 points.d Find the probability that a player got an odd number of heads, given that their score was odd.

M M Total

E 29 27

E 95 42

Total

1 2 3 4 5 6

HH

HT

TH

TT




6 Use the formula P (A ∣B) = P (A ∩ B)P (B)

to answer the following questions.

a Find P (A ∣B) if P (A ∩ B) = 0.5 and P (B) = 0.7.b Find P (A ∣B) if P (A ∩ B) = 0.15 and P (B) = 0.4.c Find P (E ∣F) if P (E ∩ F) = 0.8 and P (F) = 0.95.

7 The two events A and B in the following experiments are known to be independent.a P (A) = 0.4, and P (B) = 0.6. Find P (A ∩ B).b P (A) = 0.3, and P (B) = 0.5. Find P (A ∩ B).c P (A) = 0.4, and P (B) = 0.6. Find P (A ∣B).d P (A) = 0.7, and P (B) = 0.2. Find P (A ∣B).

8 Each of the following experiments involves two events, A and B. State in each case whether they are dependent or independent.a P (A ∣B) = 0.5 and P (A) = 0.4 and P (B) = 0.5b P (A ∣B) = 0.3 and P (A) = 0.3 and P (B) = 0.6

c P (A ∣B) = 34 and P (A) = 2

5 and P (B) = 3

10

d P (A) = 0.3 and P (B) = 0.7 and P (A ∩ B) = 0.21e P (A) = 0.2 and P (B) = 0.4 and P (A ∩ B) = 0.8

f P (A) = 12 and P (B) = 2

3 and P (A ∩ B) = 1

3

DEVELOPMENT

9 a Draw a table showing the sample space if two dice are thrown in turn and their sum is recorded.b Highlight the reduced sample space if the sum of the two dice is 5. Given that the sum of the two dice

is 5, find the probability that:i the first die shows a 1,ii at least one dice shows a 1,iii at least one of the dice shows an odd number.

10 a For two events A and B it is known that P (A ∪ B) = 0.6 and P (A) = 0.4 and P (B) = 0.3.i Use the addition formula P (A ∪ B) = P (A) + P (B) − P (A ∩ B) to find P (A ∩ B).ii Use the formula for conditional probability to find P (A ∣B).iii Find P (B ∣A).

b Suppose that P (V ∪ W) = 0.7 and P (V) = 0.5 and P (W) = 0.35. Find P (V ∣W).c Find P (X ∣Y) if P (X ∩ Y) = 0.2 and P (X) = 0.3 and P (Y) = 0.4.d Find P (A ∣B) if P (A ∪ B) = 1

3 and P (A) = 1

5 and P (B) = 3

10 .

11 Two dice are rolled. A three appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than seven.

12 The members of a cricket team know that in half of the games they played last season, they won the game and their star player Arnav was playing. Arnav consistently plays in 80% of the games. What is the probability that they will win this Saturday if Arnav is playing?




13 a A couple has two children, the older of which is a boy. What is the probability that they have two boys?

b A couple has two children and at least one is a boy. What is the probability that they have two boys?

14 A couple has three children, each being either a boy or girl.a List the sample space.b Given that at least one is a boy, what is the probability that the oldest is male?c Given that at least one of the first two children is a boy, what is the probability that the oldest is male?

15 A card is drawn from a standard pack. The dealer tells the players that it is a court card (jack, queen or king).a What is the probability that it is a jack?b What is the probability that it is either a jack or a red card?c What is the probability that the next card drawn is a jack? Assume that the first card was not replaced.

16 Two dice are tossed in turn and the outcomes recorded. Let A be the probability that the first die is odd. Let S be the probability that the sum is odd. Let M be the probability that the product is odd.a Use the definition of independence in Box 20 to find which of the three pairs of events are

independent.b Confirm your conclusions using the product-rule test for independence in Box 21.

17 The two events A and B in the following experiments are known to be independent.a P (A) = 0.4 and P (B) = 0.6. Find P (A ∪ B).b The probability of event A occurring is 0.6 and the probability of event B occuring is 0.3. What is the

probability that either A or B occurs?

18 A set of four cards contains two jacks, a queen and a king. Bob selects one card and then, without replacing it, selects another. Find the probability that:a both Bob’s cards are jacks,b at least one of Bob’s cards is a jackc given that one of Bob’s cards is a jack, the other is also a jack.

19 A small committee of two is formed from a group of 4 boys and 8 girls. If at least one of the members of the committee is a girl, what is the probability that both members are girls?

20 Jack and Ben have been tracking their success rate of converting goals in their rugby games. Jack converts 70% of his goals and Ben converts 60%. At a recent home game, both get a kick, but only one converts his goal. What is the probability that it was Ben?

21 Susan picks two notes at random from four $5, three $10 and two $20 notes. Given that at least one of the notes was $10, what is the probability that Susan has picked up a total of $20 or more?



10G456 Chapter 10 Probability

CHALLENGE

22 Researchers are investigating a potential new test for a disease, because although the usual test is totally reliable, it is painful and expensive.

The new test is intended to show a positive result when the disease is present. Unfortunately the test may show a false positive, meaning that the test result is positive, even though the disease is not present. The test may also show a false negative, meaning that the test result is negative even though the disease is in fact present.

The researchers tested a large sample of people who had symptoms that vaguely suggested that it was worth testing for the disease. They used the new test, and checked afterwards for the disease with the old reliable test, and came up with the following results:

• Of this sample group, 1% had the disease. • Of those who had the disease, 80% tested positive and 20% tested negative. • Of those who did not have the disease, 5% tested positive and 95% tested negative.

a What percentage of this group would test positive to this new test?b Use the conditional probability formula P (A ∩ B) = P (A ∣B)P (B) to find the probability that a person

tests positive, but does not have the disease.c Find the probability that the new test gives a false positive. That is, find the probability the patient

does not have the disease, given that the patient has tested positive.d Find the probability that the new test gives a false negative. That is, find the probability the patient

has the disease, given that the patient has tested negative.e Comment on the usefulness of the new test.

23 a Prove that for events A and B,

P (A ∣B) = P (B ∣A)P (B)

× P (A).

b Use this formula to produce an alternative solution to the probability of a false positive or false negative in the previous question.

24 Prove the symmetry of independence, that is, if B is independent of A then A is independent of B.

25 [A notoriously confusing question]In a television game show, the host shows the contestant three doors, only one of which conceals the prize, and the game proceeds as follows. First, the contestant chooses a door. Secondly, the host then opens one of the other two doors, showing the contestant that it is not the prize door. Thirdly, the host invites the contestant to change their choice, if they wish. Analyse the game, and advise the contestant what to do.

26 A family has two children. Given that at least one of the children is a girl who was born on a Monday, what is the probability that both children are girls?



457 Chapter 10 Review

Chapter 10 Review

Review activity • Create your own summary of this chapter on paper or in a digital document.

Chapter 10 multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF worksheet

version is also available there.

Chapter review exercise

1 If a die is rolled, find the probability that the uppermost face is:

a a two b an odd numberc a number less than two d a prime number.

2 A number is selected at random from the integers 1, 2, …, 10. Find the probability of choosing:

a the number three b an even number c a square numberd a negative number e a number less than 20 f a multiple of three.

3 From a regular pack of 52 cards, one card is drawn at random. Find the probability that the chosen card is:

a black b red c a queend the ace of spades e a club or a diamond f not a seven.

4 A student has a 63% chance of passing his driving test. What is the chance that he does not pass?

5 A fair coin is tossed twice. Find the probability that the two tosses result in:

a two tails b a head followed by a tail c a head and a tail.

6 Two dice are thrown. Find the probability of:

a a double three b a total score of fivec a total greater than nine d at least one fivee neither a four nor a five appearing f a two and a number greater than fourg the same number on both dice h a two on at least one die.

7 In a group of 60 tourists to Tasmania, 31 visited Queenstown, 33 visited Strahan and 14 visited both places. Draw a Venn diagram and find the probability that a tourist:

a visited Queenstown only b visited Strahan only c did not visit either place.

8 A die is thrown. Let A be the event that an odd number appears. Let B be the event that a number less than five appears.

a Are A and B mutually exclusive?b Find:

i P (A) ii P (B) iii P (A and B) iv P (A or B)

Revi

ew



Revi

ew

9781108469043c10_p404-458_4 Page PB 28/11/18 8:06 PM

9 The events A, B and C are independent, with P (A) = 14, and P (B) = 1

3 and P (C) = 3

5.

Use the product rule to find:

a P (AB) b P (BC) c P (AC) d P (ABC)

10 a From a standard pack of 52 cards, two cards are drawn at random without replacement. Find the probability of drawing:

i a club then a diamond, ii two hearts,iii a seven then an ace, iv the queen of hearts then the eight of diamonds.

b Repeat part a if the first card is replaced before the second card is drawn.

11 There is a 70% chance that Harold will be chosen for the boys’ debating team, and an 80% chance that Grace will be chosen for the girls’ team. Draw a probability tree diagram and find the chance that:

a Harold is chosen, but Grace is not b Grace is chosen, but Harold is notc only one of Harold and Grace is chosen d neither Harold nor Grace is chosen.

12 In a park there are four Labradors, six German shepherds and five beagles. If two dogs are selected at random, find the probability that:

a both are beagles b neither is a Labradorc at least one is a Labrador d a beagle and German shepherd are chosen.

13 There are 500 tickets sold in a raffle. The winning ticket is drawn and then the ticket for second prize is drawn, without replacing the winning ticket. If you buy 20 tickets, find the probability that you win:

a both prizes b at least one prize.

14 In a certain experiment, the probability of events A and B are P (A) = 0.6 and P (B) = 0.3 respectively. In which of these cases are the events independent?

a P (A ∩ B) = 0.18 b P (A ∣B) = 0.3 c P (A ∪ B) = 0.72

15 Two dice are rolled. A five appears on at least one of the dice. Find the probability that the sum of the uppermost faces is greater than nine.

458 Chapter 10 Probability



10web2.hunterspt-h.schools.nsw.edu.au/studentshared... · 2019-08-21 · A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is:

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